\newproblem{lay:4_6_28}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 4.6.28}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Justify the following equalities:
	\begin{enumerate}[a.]
		\item $\dim\{\mathrm{Row}\{A\}\}+\dim\{\mathrm{Nul}\{A\}\}=n$
		\item $\dim\{\mathrm{Col}\{A\}\}+\dim\{\mathrm{Nul}\{A^T\}\}=m$
	\end{enumerate}
}{
  % Solution
	\begin{enumerate}[a.]
		\item By the Rank Theorem we know that\\
		      \begin{center}
						$\dim\{\mathrm{Col}\{A\}\}+\dim\{\mathrm{Nul}\{A\}\}=n$
					\end{center}
					Since $\mathrm{Rank}\{A\}=\dim\{\mathrm{Col}\{A\}\}=\dim\{\mathrm{Row}\{A\}\}$ we have immediately the proposed equality.
		\item If we apply the Rank Theorem to $A^T$ we get
		      \begin{center}
						$\dim\{\mathrm{Col}\{A^T\}\}+\dim\{\mathrm{Nul}\{A^T\}\}=m$
					\end{center}
		      But $\dim\{\mathrm{Col}\{A^T\}\}=\dim\{\mathrm{Row}\{A\}\}=\dim\{\mathrm{Col}\{A\}\}$, and again we have the proposed equality.
	\end{enumerate}
}
\useproblem{lay:4_6_28}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
